2 Answers
2

I think you're approaching the question in the wrong way. The whole point is that you can show that for $\Re(s) > 1$,
$$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$
where the function $E(s)$ is meromorphic on the open half-plane $\Re(s) > 1/2$ with poles possibly at the zeroes of $\zeta(s)$; in fact, Zagier shows that
$$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$

Basically, this is saying that $\Phi(s)$ is meromorphic on an open neighbourhood of $\Re(s) \geq 1$ with a simple pole at $s = 1$, and the expansion above shows that the residue of $\Phi(s)$ at $s = 1$ is equal to $1$. This is precisely the same as saying that
$$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$
Indeed, we have that
$$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$
and the first limit tends to $1$ (obviously) while the second limit tends to zero (as $E(1 + \varepsilon)$ tends to something finite).

If you don't understand this method at all (i.e. all about meromorphic extensions of functions, poles, and residues), then this is probably due to a lack of background in complex analysis. Seeing as this proof of the prime number theorem is all about complex analysis, I'd recommend reading up on all these basics beforehand.

Where is it shown that $E(s)$ is holomorphic on $R(z) > 1/2$ ? To me, $E(s)$ is meromorphic on $R(z) > 1/2$, with possible poles at $s=1$ and the zeros of $\zeta(s)$. In (II), it is shown that $\zeta(s) - \frac{1}{z-1}$ extends holomorphicly to $R(s) > 0$, but I don't see how that helps $E(s)$ here.
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LowerBoundsApr 17 '11 at 2:38

Whoops, sorry. In fact what I'd written first was equivalent to the Riemann Hypothesis. It should all be fixed now. The point is that the singularities of $E(s)$ occur whenever $\zeta'(s)/\zeta(s)$ has a singularity other than at $s = 1$, because this is cancelled out by the $1/(s - 1)$ term.
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Peter HumphriesApr 17 '11 at 4:22